Spring with mass on horizontal surface

A 2.00 kg object on a horizontal frictionless surface is attached to a spring with
a spring constant 1000 N/m. The object is displaced from equilibrium 50.0 cm
horizontally and given an initial velocity of 4.0 m/s away from the equilibrium
position.

The spring was given an initial velocity + an initial displacement. So, you have factor both of these things if you are using the ##1/2kA^2## equation, remember that this is equal to the total mechanical energy in your system. It would make no sense for your spring to have a smaller amplitude if it would have more energy in the system as opposed to having no initial velocity.

Ok so how do I factor in the velocity and displacement? I thought I was with the 'v' in 0.5mv2 and the x in 0.5kx2.

What is the initial energy put into your system? Don't even worry about the amplitude yet, just figure out the total energy in the system based on the initial displacement from equilibrium and the initial velocity.